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Drupady [299]
3 years ago
11

Jim was offered a job as a computer programmer making 65K per year. Another company offered Jim a position earning $1,800 a week

. What job pays more?
Mathematics
1 answer:
steposvetlana [31]3 years ago
7 0

Answer:

since second company salary, $93,600, is more than first company yearly salary of $65000

Second company pays more

Step-by-step explanation:

To check what job pays more we have to compare yearly salary of both the company offer.

_________________________________________________

Yearly salary from one company = 65k = 65000

Weekly salary from another company = $1800

there are 52 weeks in a year

to get yearly salary , we need to multiply weekly salary with 52.

Yearly salary from another company = Weekly salary*no of wees in a year

Yearly salary from another company = 52*$1800 = $93,600

since second company salary, $93,600, is more than first company yearly salary of $65000

Second company pays more

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Help! How would I solve this trig identity?
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Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}

Then the identity can be rewritten as:

sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\

Now we can multiply both sides by cos⁴(x) to get:

\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)

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sin²(x) + cos²(x) = 1

1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

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2 years ago
A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.
Reptile [31]

Answer:

a.P.I=\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)

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Step-by-step explanation:

We are given that a linear differential equation

y''+2y=e^{3x}+x^3

We have to find the particular solution

P.I=\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}

P.I=\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}

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P.I=\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0} (higher order terms can be neglected

P.I=\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)

b.Characteristics equation

D^2+2=0

D=\pm\sqrt2 i

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G.S=C.F+P.I

G.S=C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)

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